Value of $ \lim_{n \to \infty} \int \limits_{0}^{1}nx^n e^{ x^2} ?$ How to find the value of $$ \lim_{n \to \infty}\int \limits_{0}^{1} nx^n e^{ x^2} ?$$
From wolfram the limit approaches to $e$ for larger values of $n$. I substituted $x^2 $ with $u$ and obtained 
$$ \frac{ n} {2} \int \limits_{0}^{1} u^{\frac{n-1}{2}} e^{u} du $$ 
The value of this integral can be obtained from  here. But still I'm unable to get it. Is there any better approach for this question?
 A: Let 
$$I_n=\int_0^1x^ne^{x^2}dx$$
$I_n$ is decreasing because:
$$I_n-I_{n+1}=\int_0^1x^n(1-x)e^{x^2}dx\geq 0$$
We want to find $\lim\limits_{n\to \infty} nI_n$. Integrating by parts, we can see that:
$$I_n+\frac{2}{n+1}I_{n+2}=\frac{e}{n+1}$$
Now, since $I_n\geq I_{n+2}$, we have:
$$\frac{e}{n+1}\leq I_n+\frac{2}{n+1}I_n\Rightarrow I_n \geq \frac{e}{n+3}$$
Also because $I_n\leq I_{n-2}$, we get
$$\frac{e}{n-1}=I_{n-2}+\frac{2}{n-1}I_n\geq I_n+\frac{2}{n-1}I_n\Rightarrow I_n \leq \frac{e}{n+1}$$
Chaining these inequalities together, we have:
$$\frac{e}{n+3}\leq I_n \leq \frac{e}{n+1}$$
or
$$\frac{n}{n+3}e\leq nI_n\leq \frac{n}{n+1}e$$
Squeezing, we conclude that:
$$\lim\limits_{n\to \infty} nI_n = e$$
A: You have the right idea about changing the variable, just a different change: $u=x^{n+1}$
$$
\begin{align}
\lim_{n\to\infty}\int_0^1nx^ne^{x^2}\,\mathrm{d}x
&=\lim_{n\to\infty}\int_0^1\frac{n}{n+1}e^{u^{\frac2{n+1}}}\,\mathrm{d}u\\
&=\int_0^11\cdot e^1\,\mathrm{d}u\\[6pt]
&=e
\end{align}
$$
Note that $\frac{n}{n+1}e^{u^{\frac2{n+1}}}$ increases monotonically to $e$ for all $u\in(0,1]$, and uniformly on compact subsets, so we can use monotone convergence, dominated convergence, or uniform convergence (on each compact subset) to justify the exchange of limit and integral.
A: You can get an upper bound because for $0\le x\le1$, $e^{x^2}\le e$ so
$$\int_0^1nx^ne^{x^2}dx\le\frac{ne}{n+1}$$
We know that for $1-\epsilon\le x\le1$, $e^{x^2}\ge e^{1-2\epsilon+\epsilon^2}\ge e\cdot e^{-2\epsilon}\ge e(1-2\epsilon)$ because $e^{-x}\ge(1-x)$ for $x\ge0$, the latter function begin the linearization of the former at $x=0$ and the former being concave up. Also
$$1-(1-\epsilon)^{n+1}\ge1-e^{-\epsilon(n+1)}$$
So a lower bound is
$$\begin{align}\int_0^1nx^ne^{x^2}dx&\ge\int_{1-n^{-1/2}}^1nx^ne^{x^2}dx\ge\int_{1-n^{-1/2}}^1nx^ne(1-2n^{-1/2})dx\\
&=\frac{ne(1-2n^{-1/2})}{n+1}\left(1-(1-n^{-1/2})^{n+1}\right)\\
&\ge\frac{ne(1-2n^{-1/2})}{n+1}\left(1-e^{-n^{-1/2}(n+1)}\right)\end{align}$$
Since our integral lies between two quantities that approach $e$ as $n\rightarrow\infty$, the linit is in fact $e$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With
Laplace Method:
\begin{align}
\lim_{n \to \infty}\int_{0}^{1}nx^{n}\expo{x^{2}}\dd x & =
\lim_{n \to \infty}\bracks{n\int_{0}^{1}
\pars{1 - x}^{n}\expo{\pars{1 - x}^{2}}\dd x}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n\int_{0}^{1}
\expo{n\ln\pars{1 - x}}\expo{\pars{1 - x}^{2}}\dd x}
\\[5mm] & =
\lim_{n \to \infty}\bracks{n\int_{0}^{\infty}
\expo{-nx}\expo{\pars{1 - 0}^{2}}\dd x} =
\bbx{\Large \expo{}}
\end{align}
